TheAlgorithms-C-Plus-Plus/Backtracking/Rat_maze.cpp

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2018-03-13 07:06:21 +08:00
/*
A Maze is given as N*N binary matrix of blocks where source block is the upper
left most block i.e., maze[0][0] and destination block is lower rightmost
block i.e., maze[N-1][N-1]. A rat starts from source and has to reach destination.
The rat can move only in two directions: forward and down. In the maze matrix,
0 means the block is dead end and 1 means the block can be used in the path
from source to destination.
*/
#include <iostream>
#define size 4
using namespace std;
int solveMaze(int currposrow,int currposcol,int maze[size][size],int soln[size][size])
{
if((currposrow==size-1) && (currposcol==size-1))
{
soln[currposrow][currposcol]=1;
for(int i=0;i<size;++i)
{
for (int j=0;j<size;++j)
{
cout<<soln[i][j];
}
cout<<endl;
}
return 1;
}
else
{
soln[currposrow][currposcol]=1;
// if there exist a solution by moving one step ahead in a collumn
if((currposcol<size-1) && maze[currposrow][currposcol+1]==1 && solveMaze(currposrow,currposcol+1,maze,soln))
{
return 1;
}
// if there exists a solution by moving one step ahead in a row
if((currposrow<size-1) && maze[currposrow+1][currposcol]==1 && solveMaze(currposrow+1,currposcol,maze,soln))
{
return 1;
}
// the backtracking part
soln[currposrow][currposcol]=0;
return 0;
}
}
int main(int argc, char const *argv[])
{
int maze[size][size]={
{1,0,1,0},
{1,0,1,1},
{1,0,0,1},
{1,1,1,1}
};
int soln[size][size];
for(int i=0;i<size;++i)
{
for (int j=0;j<size;++j)
{
soln[i][j]=0;
}
}
int currposrow=0;
int currposcol=0;
solveMaze(currposrow,currposcol,maze,soln);
return 0;
}